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(B) The kinetic energy $(KE)$ of a simple harmonic oscillator is given by $KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$ and the total energy $(TE)$ is $TE

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$\frac{1}{3} \ s$

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(B) The kinetic energy $(KE)$ of a simple harmonic oscillator is given by $KE = \frac{1}{2} m \omega^2 (A^2 - x^2)$ and the total energy $(TE)$ is $TE = \frac{1}{2} m \omega^2 A^2$. Given that $KE = 75\% 	ext{ of } TE$, we have: $KE = \frac{3}{4} TE$ $\frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{3}{4} (\frac{1}{2} m \omega^2 A^2)$ $A^2 - x^2 = \frac{3}{4} A^2$ $x^2 = A^2 - \frac{3}{4} A^2 = \frac{1}{4} A^2$ $x = \pm \frac{A}{2}$ For a particle starting from the mean position ($x=0$ at $t=0$), the displacement is given by $x = A \sin(\omega t)$. Setting $x = \frac{A}{2}$, we get $\frac{A}{2} = A \sin(\omega t) \Rightarrow \sin(\omega t) = \frac{1}{2}$. This implies $\omega t = \frac{\pi}{6}$. Since $\omega = \frac{2\pi}{T}$, we have $\frac{2\pi}{T} t = \frac{\pi}{6}$. $t = \frac{T}{12}$. Given $T = 4 \ s$, $t = \frac{4}{12} = \frac{1}{3} \ s$.

$A$ body starting at $t=0$ from the origin oscillates simple harmonically with a period of $4 \ s$. After what time will its kinetic energy be $75 \%$ of its total energy?

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